Let f x = x + log e ⁡ x - x log e ⁡ x , x ∈ 0 ,   ∞ • Column 1 contains…

$f\left(x\right)=x+\ell nx-x\ell nx, x>0$

$f^{\text{'}}\left(x\right)=1+\frac{1}{x}-\ell nx-1$

$f^{\text{'}\text{'}}\left(x\right)= -\frac{1}{x^2}-\frac{1}{x}=\frac{-\left(x+1\right)}{x^2}$

In Column 1,

(I)  $f\left(1\right) f\left(e^2\right)<0$ so true

(II)   $f^{\text{'}}\left(1\right) f^{\text{'}}\left(e\right)<0$ so true

(III)   $f^{\text{'}}\left(x\right)$ is always positive for $(0,1)$ so it cannot be zero. So (III) is false.

(IV) Is false

In Column 2

As   $\underset{x\to \infty }{\lim }f\left(x\right)= \underset{x\to \infty }{\lim }x\left[1+\frac{\ell n x}{x}-\mathcal{ }\ell n x\right]= -\infty$

$\therefore$ (i)  is false (ii) is true

$\underset{x\to \infty }{\lim }{f}^{\text{'}}\left(x\right)= -\infty$ ,so (iii) is true.

$\underset{x\to \infty }{\lim }{f}^{\text{'}\text{'}}\left(x\right)=0$, so (iv) is true.

(P)  $f^{\text{'}}\left(x\right)$  is positive in $(0, 1)$,so true.

(Q)  $f^{\text{'}}\left(x\right)<0$ ,for in $\left(e, e^2\right)$ ,so true.

As  $f^{\mathrm{\text{'}\text{'}}}\left(x\right)<0 \forall x>0$ therefore, R is false, S is true.

Alternate Solution:

$f\left(x\right)=x+ \ell n x-x\ell n x$

$f^{\text{'}}\left(x\right)=\frac{1}{x}-\ell n x=0$ at $x=x_0$ where $x_0\in \left(1, e\right)$

$f^{\text{'}\text{'}}\left(x\right)= -\frac{1}{x^2}-\frac{1}{x}<0 \forall x>0 \Rightarrow f\left(x\right)$  concave down